A tank that resembles a hemisphere flat face up is full of water, with a small outlet on the top right. The weight of water is 62.5 lb/ft^3, find the work required to pump the water out of the outlet.

To find the work required to pump the water out of a hemispherical tank with a small outlet at the top, we need to use the concept of work in the context of pumping liquids. The work \( W \) is calculated by integrating the force exerted over a distance. Here's how we can solve this step-by-step:

### 1. **Define the Problem:**
- **Shape of the Tank:** Hemisphere with the flat face up.
- **Weight of Water:** \( 62.5 \) lb/ft³.
- **Radius of the Hemisphere:** Let’s denote it as \( R \).

### 2. **Determine the Differential Element:**
Consider a thin horizontal slice of water at a height \( y \) from the bottom of the tank, where the bottom of the hemisphere is at \( y = 0 \) and the top of the hemisphere is at \( y = R \).

- **Thickness of the slice:** \( dy \).
- **Volume of the slice:** This can be represented as the area of the circular cross-section at height \( y \) times the thickness \( dy \).

  The area of the cross-section is given by the formula for the area of a circle with radius \( \sqrt{R^2 - y^2} \) (since the radius at height \( y \) decreases as we move up).

  \[
  \text{Area} = \pi \left( R^2 - y^2 \right)
  \]

  So, the volume \( dV \) of the slice is:

  \[
  dV = \pi (R^2 - y^2) \, dy
  \]

### 3. **Calculate the Weight of the Slice:**
The weight \( dW \) of this thin slice is given by the density of water times the volume:

\[
dW = 62.5 \cdot \pi (R^2 - y^2) \, dy
\]

### 4. **Calculate the Work to Pump this Slice to the Top:**
To find the work required to pump this slice out of the outlet at the top, we need to move it a distance of \( R - y \). The work \( dW \) to move the slice is given by:

\[
d\text{Work} = dW \cdot (R - y) = 62.5 \cdot \pi (R^2 - y^2) (R - y) \, dy
\]

### 5. **Integrate Over the Entire Volume:**
To find the total work, we integrate from \( y = 0 \) to \( y = R \):

\[
W = \int_0^R 62.5 \cdot \pi (R^2 - y^2) (R - y) \, dy
\]

### 6. **Evaluate the Integral:**
Simplifying the integral:

\[
W = 62.5 \pi \int_0^R (R^3 - R y^2 - R^2 y + y^3) \, dy
\]

Let's break it down:

\[
\int_0^R (R^3 - R y^2 - R^2 y + y^3) \, dy = \left[ R^3 y - \frac{R}{3} y^3 - \frac{R^2}{2} y^2 + \frac{y^4}{4} \right]_0^R
\]

Substitute \( y = R \) and \( y = 0 \):

\[
R^3(R) - \frac{R}{3} (R)^3 - \frac{R^2}{2} (R)^2 + \frac{(R)^4}{4}
\]
\[
= R^4 - \frac{R^4}{3} - \frac{R^4}{2} + \frac{R^4}{4}
\]
Combine like terms:

\[
R^4 \left(1 - \frac{1}{3} - \frac{1}{2} + \frac{1}{4}\right) = R^4 \left(\frac{12}{12} - \frac{4}{12} - \frac{6}{12} + \frac{3}{12}\right)
\]
\[
= R^4 \left(\frac{5}{12}\right)
\]

Finally, multiply by the constant:

\[
W = 62.5 \pi R^4 \cdot \frac{5}{12} = \frac{62.5 \pi R^4}{2.4}
\]

So, the total work required is:

\[
W = \frac{312.5 \pi R^4}{12} \text{ lb-ft}
\]

This gives the amount of work needed to pump all the water out of the hemisphere.

### Questions to Consider:

1. How would the work change if the outlet was at the center of the flat face instead of the top right?
2. What happens to the work required if the tank was not full?
3. How would this problem differ if the tank was a full sphere?
4. What if the density of the fluid was different?
5. How does the shape of the tank influence the required work?

### Tip:
Always break down the integral into simpler parts to avoid mistakes in complex expressions!

Learn how to calculate the work required to pump water from a hemispherical tank with a small outlet at the top. This problem involves calculus concepts like integration and volume of revolution, and is suitable for university-level students studying physics and engineering.

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